Notes on math, coding, and other stuff

Tag: cayley

Today me and a few of my friends took the Cayley math contest. This is another University of Waterloo contest with 25 multiple choice problems and a time limit of 60 minutes.

I did not get the last problem, and did the second last problem incorrectly. Probably no medal for me, even if I answered everything else correctly.

I think the most challenging problem on this contest was the last one:

The numbers 0 to 9 are arranged in a circle in order (like a clock). Steve places a counter at 0. On his first move, he moves the counter step clockwise (to 1). On the second move he moves it steps (to 5). If this continues, what position will the counter be after 1234 moves?

My solution

It’s funny that I had thought there were 12 numbers on the circle, where there was actually 10. Probably because there was already a clock related problem earlier in the contest. It would be slightly easier with 10.

The diagram in the booklet looked something like this:

The problem is needlessly confusing. It really just wants you to find the value of this expression:

So we’re basically looking for the last digit of the expression. We just have to keep track of the units digit as we go along.

In the list [1..1234] there are 123 numbers ending with 0, 123 ending with 5, 123 ending with 6, and so on. Also there are 124 ending with 1, 2, 3, and 4.

This post was inspired by a problem I came across when studying for a math contest, namely problem 23 in the 2007 Cayley Contest.

I think this is an interesting problem because the solution is not so obvious, and there are several incorrect approaches that I’ve tried (and given up on). I’ve modified the problem slightly to make it more interesting.

Here we have a rectangle, . It’s rotated to the right by degrees, using as the pivot. This forms as the new rectangle. We are given and , we are asked to find the area of the shaded region.

Can you come up with a general formula for the area of the shaded region, using the given variables , , and ?

Solution

I will use the notation to denote the area of triangle .

The solution is to draw a line parallel to through , the blue line. This splits the lower shaded area into two triangles: and , and a rectangle: .

Since is the same as , the top shaded area is the same as the bottom shaded area.

Let’s find the area of triangle .

The angle is equal to . Of course, is equal to . Using some basic trigonometry, , , and . Now we know the area of triangle :

Next we find the area of . , and using trigonometry again, .

Finally, for the rectangle :

Now we add all this up and simplify ( is total):

It’s rather tedious to get from the second to the last step, but Wolfram Alpha could do it for me.

Notice that we assume that intersects . If is large enough, then would intersect , and this formula would no longer work.

Of course it would also fail if is greater than , so my formula works for only a rather limited range of values.